Question

For the reaction \( \text{2N}_2\text{O}_5 \rightarrow 4\text{NO}_2 (g) + \text{O}_2 (g) \), the initial concentration of \( \text{N}_2\text{O}_5 \) is 2.0 mol L\(^{-1}\), and after 300 minutes, it is reduced to 1.4 mol L\(^{-1}\). The rate of production of \( \text{NO}_2 \) (in mol L\(^{-1}\) min\(^{-1}\)) is

• A. \( 4 \times 10^{-4} \)
• B. \( 2.5 \times 10^{-3} \)
• C. \( 4 \times 10^{-3} \)
• D. \( 2.5 \times 10^{-4} \)

Hint:

For reactions with stoichiometric relationships, the rate of formation or consumption of a product or reactant can be calculated using the rate of change of one species and the stoichiometric coefficients.

Answer 1

The Correct Option is C

The rate of production of \( \text{NO}_2 \) can be determined from the rate of change of \( \text{N}_2\text{O}_5 \). The balanced equation shows that for every 2 moles of \( \text{N}_2\text{O}_5 \) consumed, 4 moles of \( \text{NO}_2 \) are produced. Therefore, the rate of production of \( \text{NO}_2 \) is: \[ \frac{d[\text{NO}_2]}{dt} = \frac{4}{2} \times \frac{d[\text{N}_2\text{O}_5]}{dt} \] Using the concentration change of \( \text{N}_2\text{O}_5 \): \[ \Delta [\text{N}_2\text{O}_5] = 2.0 - 1.4 = 0.6 \, \text{mol L}^{-1} \] Since the reaction occurs over 300 minutes, the rate of change of \( \text{N}_2\text{O}_5 \) is: \[ \frac{d[\text{N}_2\text{O}_5]}{dt} = \frac{0.6}{300} = 2 \times 10^{-3} \, \text{mol L}^{-1} \text{min}^{-1} \] Thus, the rate of production of \( \text{NO}_2 \) is: \[ \frac{d[\text{NO}_2]}{dt} = 4 \times 10^{-3} \, \text{mol L}^{-1} \text{min}^{-1} \]